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I was struck by the fact that Classic Gauge Theory in EM says that E & B fields are invariant under transformations of the potential, $V$, and vector potential, $A$, by any function, $f$, that is twice continuously differentiable function that depends on position and time. That is, \begin{align} \mathbf{A} &\mapsto \mathbf{A} + \nabla f\\ V &\mapsto V - \frac{\partial f}{\partial t} \end{align}

Which is remarkably similar to the invariance of action integral under the addition of the total derivative with respect to $t$ of any function of position and time.

I wanted to dig more into this, and I saw that gradient in spherical has the same terms as the square rooted the dr-dot squared portion of the Lagranian (ie the kinetic) and divided by the non-dotted term. It's hard to put what I mean into words. However, please refer to the comparison below.

Gradient: $$\boldsymbol{\nabla}{f} = \mathbf{e}_r\, \frac{\partial{f}}{\partial{r}} + \mathbf{e}_\theta\,\frac{1}{r } \,\frac{\partial{f}}{\partial{\theta}} + \mathbf{e}_\phi\,\frac{1}{r\,\sin{\theta}} \,\frac{\partial{f}}{\partial{\phi}}$$ Lagrangian: $$\vec v^2 = \dot r^2 + r^2 \dot \theta^2 + r^2 \sin^2(\theta) \dot \phi^2$$

Am I grasping at straws with a connection between Gauge invariance and Action invariance? Am I grasping at straws for connection between gradient and kinetic?

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The equations of motion of the Lagrangian of electrodynamics are gauge invariant under the gauge transformation $A^\mu = A^\mu + \partial^u f$ where $A^0=V$.

You are seeing similarities in these expressions because of how vectors and basis vectors transform into spherical coordinates and/or from the metric of spherical coordinates, depending on how you are approaching the question.

Consider polar coordinates.

We want to express $v=\frac{dx}{dt} i + \frac{dy}{dt}j$ in polar coordinates.

$x=r \cos(\theta)$

$y=r \sin(\theta)$

Use the Chain Rule and Product Rule.

$\dot{x}= \dot{r} \cos{\theta} -r \sin(\theta) \dot{\theta}$

$\dot{y}= \dot{r} \sin{\theta} +r \cos(\theta) \dot{\theta}$

$v=(\dot{r} \cos{\theta} -r \sin(\theta) \dot{\theta})i + (\dot{r} \sin{\theta} +r \cos(\theta) \dot{\theta})j $

$\text{velocity}=(\dot{x},\dot{y})=\dot{x} \hat{i}+\dot{y} \hat{j}=f(r,\theta) \hat{r} + g(r,\theta) \hat{\theta}$

Use the fact that basis vectors transform like covariant tensors of rank 1.

$\hat{r}= \frac{dx}{dr} \hat{i} + \frac{dy}{dr} \hat{j} \\$

$\hat{\theta}= \frac{dx}{d\theta} \hat{i}+\frac{dy}{d\theta} \hat{j} $

And

$\hat{r} \cdot \text{velocity} = f(r,\theta)$

$\hat{\theta} \cdot \text{velocity} = g(r,\theta)$.

Therefore,

$f(r,\theta)=\cos(\theta)[(\dot{r} \cos{\theta} -r \sin(\theta) \dot{\theta}]+ \sin(\theta)[(\dot{r} \sin{\theta} +r \cos(\theta) \dot{\theta})] = \dot{r}$

(using $\cos^2 (\theta) + \sin^2 (\theta)=1)$

The $\theta$ component of velocity can be found similarly.

This ultimately comes from the fact that distance is an invariant such that

$dx^2 + dy^2 =(\frac{dx}{dr} \frac{dx}{dr} +\frac{dy}{dr} \frac{dy}{dr})dr^2 +(\frac{dx}{d\theta} \frac{dx}{d\theta} +\frac{dy}{d\theta} \frac{dy}{d\theta})d\theta^2 = \hat{r} \cdot \hat{r} \space dr^2 + \hat{\theta} \cdot \hat{\theta} d\theta^2 $

The mathematics for the gradient is similar. The gradient is a covariant vector and the velocity vector here is a contravariant vector. These transform "oppositely" to each other and that's why things are moving around from the denominator to the numerator.

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  • $\begingroup$ Thank you! I'll return to think when I've got better understanding of tensor calc which we just started in EM $\endgroup$
    – EEH
    Oct 28, 2022 at 20:18

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